Please Explain (actually explain) The Monty Hall Problem

[sbrothy]

Nice thread, I do not understand all of the explanations, especially the entropy one! (wow!)

I have an explanation of why you should always switch, it would be good if the maths guys gave a view, I would say it is more of a qualitative/ intuitive explanation, although there are some numbers in there.

At the beginning of the game you are more likely to pick a goat simply because there are more goats than cars, only 33% chance of a car. So the car is always more likely to be in the other 67% that you did not pick i.e. one of the other 2 doors.

So when half of that 67% is revealed as goat, it is then logical to switch from your door to that other door.

I can't say I follow all of them either. Especially the Bayesian one.

I suspect Claude Shannon would have considered the matter decided by now though.

 

[WWGD]

Just to add to what was said, consider the probability associated with the option of switching vs staying. Event car behind door displayed ( but not shown) by Monty, is the only undesirable option.

 

[jbergman]

I want to elaborate on my previous post. Imagine the game is changed.

You have three doors, one, with a car behind it. Now you can either choose to look behind 2 doors or just 1. Obviously everyone will look behind 2.

This is identical to the Month Hall problem!

For example, if you would pick doors 1 and 3 in this new game, you can do the exact same in the original Monty Hall game!

Just choose door 2, and then the host will open one of doors 1 or 3 and then you can switch to open the other door.

We don't really need to use probability. It's plainly obvious that anyone would rather get the chance to choose 2 doors instead of 1.

 

[Dale]

In that simulation/illustration, what did you devise/implement for/as the algorithmic/procedural basis by which the simulated player uses information from prior games cumulatively to discern/evince the rule?

Oops, sorry I missed this earlier.

This is a Bayesian updating approach. I used a uniform prior which is a conjugate prior. So after observing $w$ wins out of $g$ games for a given strategy $\text{door}=d,\ \text{switch}=s$ the posterior is a beta distribution $p(d,s) \sim \beta(1+w,1+g-w)$. The uniform distribution (the prior) is $\beta(1,1)$. Then the probability of inferring the correct strategy is the probability that $p(d_1,s_1)>p(d_2,s_2)$.